Question

With unsigned char you can store a number from 0 to 255

255(b10) = 11111111(b2)             <=    that’s 1 byte

This will make it easy to preform operations like +,-,*…

Now how about:

255(b10) = 10101101(b2)

Following this method will make it possible to represent up to 399 using unsigned char?

399(b10) = 11111111(b2)

Can someone propose an algorithm to preform addition using the last method?

Answer 1

With eight bits there are only 256 possible value (2⁸), no matter how you slice and dice it.

Your scheme to encode digits in a 2-3-3 form like:

255 = 10 101 101
399 = 11 111 111

ignores the fact that those three-bit sequences in there can only represent eight values (0-7), not ten (ie, that second one would be 377, not 399).

The trade-off is that this means you gain the numbers '25[6-7]' (2 values) '2[6-7][0-7]' (16 values) and '3[0-7][0-7]' (64 values) for a total of 82 values.

Your sacrifice for that gain is that you can no longer represent any numbers containing 8 or 9: '[8-9]' (2 values), '[1-7][8-9]' (14 values), '[8-9][0-9]' (20 values), '1[0-7][8-9]' (16 values), '1[8-9][0-9]' (20 values) or '2[0-4][8-9]' (10 values), for a total of 82 values.

The balance there (82 vs. 82) shows that there are still only 256 possible values for an eight-bit data type.

So your encoding scheme is based on a flawed premise, which makes the second part of your question (how to add them) irrelevant, I’m afraid.